1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Arte-miy333 [17]
3 years ago
15

A posição de um veículo no trecho reto de uma estrada, durante um certo tempo e a partir de um estabelecido referencial, tem a s

eguinte forma: S = – 4,0 + 40t (no MKS). Pode-se afirmar que:
a) Seu movimento é retrógrado.



b) Sua aceleração centrípeta é nula.



c) Ele inverte o sentido de movimento em t = 16 s.



d) Ele passa na origem dos espaços no instante t = 80 s.
Physics
1 answer:
Mariulka [41]3 years ago
4 0

Answer:

b) Sua aceleração centrípeta é nula.

Explanation:

a)

Movimento Retrógrado é o movimento descrito por um móvel cujo movimento vai  percorrendo em sentido contrário ao da reta.

É descrito matematicamente por uma função linear <u>decrescente</u>.

Analisando a função abaixo, percebe-se que esse movimento não é retrógrado. Haja vista, ser esta função crescente uma vez que o sinal da constante "v" é positivo.

S=-4+40t

Vamos testar para verificar

S=-4+4(1) \therefore S=0\\S=-4+40(2) \therefore S=-4+80 \therefore S=76 m\\S= -4 +40(3) \therefore S=-4+120 \therefore S=116 metros

b)

A Aceleração centrípeta está diretamente ligada ao Movimento Circular, podemos calculá-la por

a_{cp}=\frac{v^{2}}{R}

Como a função S=-4+4t descreve um movimento linear, sua aceleração centrípeta é nula, pois não há Movimento Circular sendo executado pelo móvel. Assim, não há Raio

Logo:

c)  Para que houvesse inversão de sentido, o valor de S teria que ser negativo. Fazendo a substituição vemos que isso não ocorre:

S=-4+40*16\\S=-4+640\\S=636 m

You might be interested in
Which of the following is a correct equation for total energy?
Liula [17]

Answer:

kinetic energy + potential energy

6 0
3 years ago
Read 2 more answers
a ball is thrown inclined in to the air with a initial velocity of u. if it reaches the maximum height in 6 seconds , find the r
olga55 [171]

Answer:

11:1

Explanation:

At constant acceleration, an object's position is:

y = y₀ + v₀ t + ½ at²

Given y₀ = 0, v₀ = u, and a = -g:

y = u t − ½g t²

After 6 seconds, the ball reaches the maximum height (v = 0).

v = at + v₀

0 = (-g)(6) + u

u = 6g

Substituting:

y = 6g t − ½g t²

The displacement between t=0 and t=1 is:

Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]

Δy = 6g − ½g

Δy = 5½g

The displacement between t=6 and t=7 is:

Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]

Δy = (42g − 24½g) − (36g − 18g)

Δy = 17½g − 18g

Δy = -½g

So the ratio of the distances traveled is:

(5½g) / (½g)

11 / 1

The ratio is 11:1.

8 0
4 years ago
Scientific theories are explanations for naturally occurring events or phenomena. Although scientific theories require extensive
Arada [10]
The answer should be B) Scientific theories and laws develop from the acquisition of scientific knowledge. Hope this helps you.
3 0
3 years ago
Read 2 more answers
1. ¿Cuál es el actor clave de del exceso de peso corporal?
Slav-nsk [51]

El factor mas importante para el exceso de peso es un  exceso de energía creada por una alimentación excesiva

El peso de un cuerpo es definido por la relación entre la energía requerida para los procesos vitales del cuerpo, sus actividades físicas diarias y la energía suministra en forma de alimentos.

Cuando estos dos parámetros están en balance el peso es estable, pero cuando la cantidad de alimentos aumenta o el valor energético de los mismo aumenta se tiene un exceso de energía que el cuerpo almacena en forma de grasa corporal, este el el factor mas importante para el exceso de peso.

En conclusión el factor mas importante para el exceso de peso es un  exceso de energía por una alimentación excesiva

aprende mas acerca del peso corporal aquí:

brainly.com/question/13032223

8 0
3 years ago
A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region
zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with  due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

K_{1} = K_{2} + W_{f}

Where:

K_{1}, K_{2} are the initial and final translational kinetic energies of the tobbogan, measured in joules.

W_{f} - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})

Where:

f - Friction force, measured in newtons.

\Delta s - Distance travelled by the toboggan in the rough region, measured in meters.

m - Mass of the toboggan, measured in kilograms.

v_{1}, v_{2} - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}

If m = 375\,kg, v_{1} = 4.50\,\frac{m}{s}, v_{2} = 1.20\,\frac{m}{s} and \Delta s = 5.40 \,m, then:

f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}

f = 653.125\,N

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%

\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%

\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%

\%K_{loss} = 92.889\,\%

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%

\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%

\%v_{loss} = 73.333\,\%

The speed of the toboggan is reduced in 73.333 %.

5 0
3 years ago
Other questions:
  • Explain the main difference between mixtures and compounds
    9·1 answer
  • A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the ma
    5·1 answer
  • Can someone help me with this. I'm not really sure if the right answer is c.
    10·1 answer
  • What is the SI unit for speed and velocity?
    14·2 answers
  • A car traveling 90km/hr is 100 m behind a truck traveling 50km/hr. How long will it take the car to reach the truck?
    15·2 answers
  • You are riding in a hot air balloon that, relative to the ground, has a velocity of 6.0 m/s due east. You see a hawk moving dire
    9·1 answer
  • Gizmo Warm-up: Lifting a piano A pulley is a simple machine that is used to lift heavy objects. A pulley is a wheel with a groov
    10·1 answer
  • An organism that cannot make its own food called
    11·2 answers
  • If an object with a mass of 248kg is lifted 76 meters in 18 seconds, how much power is used?
    6·1 answer
  • A proton is projected in the positive x direction into a region of a uniform electric field →E =(-6.00 × 10⁵) i^ N/C at t=0 . Th
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!